# Nontransitive dice

(Redirected from Efron's dice)

A set of dice is nontransitive if it contains three dice, A, B, and C, with the property that A rolls higher than B more than half the time, and B rolls higher than C more than half the time, but it's not true that A rolls higher than C more than half the time. In other words, a set of dice is nontransitive if its "rolls a higher number than more than half the time" relation is not transitive.

It is possible to find sets of dice with the even stronger property that, for each die in the set, there is another die that rolls a higher number than it more than half the time. Using such a set of dice, one can invent games which are biased in ways that people unused to nontransitive dice might not expect (see Example).

## Example

An example of nontransitive dice (opposite sides have the same value as those shown).

Consider the following set of dice.

• Die A has sides 2, 2, 4, 4, 9, 9.
• Die B has sides 1, 1, 6, 6, 8, 8.
• Die C has sides 3, 3, 5, 5, 7, 7.

The probability that A rolls a higher number than B, the probability that B rolls higher than C, and the probability that C rolls higher than A are all 5/9, so this set of dice is nontransitive. In fact, it has the even stronger property that, for each die in the set, there is another die that rolls a higher number than it more than half the time.

Now, consider the following game, which is played with a set of dice.

1. The first player chooses a die from the set.
2. The second player chooses one of the remaining dice.
3. Both players roll their dice; the player who rolls the higher number wins.

If this game is played with a transitive set of dice, it is either fair or biased in favor of the first player, because the first player can always find a die that will not be beaten by any other die more than half the time. If it is played with the set of dice described above, however, the game is biased in favor of the second player, because the second player can always find a die that will beat the first player's die with probability 5/9.

## Variations of nontransitive dice

### Efron's dice

Efron's dice are a set of four nontransitive dice invented by Bradley Efron.

Efron's dice.

The four dice A, B, C, D have the following numbers on their six faces:

• A: 4, 4, 4, 4, 0, 0
• B: 3, 3, 3, 3, 3, 3
• C: 6, 6, 2, 2, 2, 2
• D: 5, 5, 5, 1, 1, 1

#### Probabilities

Each die is beaten by the previous die in the list, with a probability of 2/3:

$P(A>B) = P(B>C) = P(C>D) = P(D>A) = {2 \over 3}$
A conditional probability tree can be used to discern the probability with which C rolls higher than D.

B's value is constant; A beats it on 2/3 rolls because four of its six faces are higher.

Similarly, B beats C with a 2/3 probability because only two of C's faces are higher.

P(C>D) can be calculated by summing conditional probabilities for two events:

• C rolls 6 (probability 1/3); wins regardless of D (probability 1)
• C rolls 2 (probability 2/3); wins only if D rolls 1 (probability 1/2)

The total probability of win for C is therefore

$\left( {1 \over 3}\times1 \right) + \left( {2 \over 3}\times{1 \over 2} \right) = {2 \over 3}$

With a similar calculation, the probability of D winning over A is

$\left( {1 \over 2}\times1 \right) + \left( {1 \over 2}\times{1 \over 3} \right) = {2 \over 3}$

#### Best overall die

The four dice have unequal probabilities of beating a die chosen at random from the remaining three:

As proven above, die A beats B two-thirds of the time but beats D only one-third of the time. The probability of die A beating C is 4/9 (A must roll 4 and C must roll 2). So the likelihood of A beating any other randomly selected die is:

${1 \over 3}\times \left( {2 \over 3} + {1 \over 3} + {4 \over 9} \right) = {13 \over 27}$

Similarly, die B beats C two-thirds of the time but beats A only one-third of the time. The probability of die B beating D is 1/2 (only when D rolls 1). So the likelihood of B beating any other randomly selected die is:

${1 \over 3}\times \left( {2 \over 3} + {1 \over 3} + {1 \over 2} \right) = {1 \over 2}$

Die C beats D two-thirds of the time but beats B only one-third of the time. The probability of die C beating A is 5/9. So the likelihood of C beating any other randomly selected die is:

${1 \over 3}\times \left( {2 \over 3} + {1 \over 3} + {5 \over 9} \right) = {14 \over 27}$

Finally, die D beats A two-thirds of the time but beats C only one-third of the time. The probability of die D beating B is 1/2 (only when D rolls 5). So the likelihood of D beating any other randomly selected die is:

${1 \over 3}\times \left( {2 \over 3} + {1 \over 3} + {1 \over 2} \right) = {1 \over 2}$

Therefore the best overall die is C with a probability of winning of 0.5185. C also rolls the highest average number in absolute terms, 3 13. (A's average is 2 23, while B's and D's are both exactly 3.)

#### Variants with equal averages

Note that Efron's dice have different average rolls: the average roll of A is 8/3, while B and D each average 9/3, and C averages 10/3. The non-transitive property depends on which faces are larger or smaller, but does not depend on the absolute magnitude of the faces. Hence one can find variants of Efron's dice where the odds of winning are unchanged, but all the dice have the same average roll. For example,

• A: 6, 6, 6, 6, 0, 0
• B: 4, 4, 4, 4, 4, 4
• C: 8, 8, 2, 2, 2, 2
• D: 7, 7, 7, 1, 1, 1

or

• A: 7, 7, 7, 7, 1, 1
• B: 5, 5, 5, 5, 5, 5
• C: 9, 9, 3, 3, 3, 3
• D: 8, 8, 8, 2, 2, 2

These variant dice are useful, e.g., to introduce students to different ways of comparing random variables (and how only comparing averages may overlook essential details).

### Numbered 1 through 24 dice

A set of four dice using all of the numbers 1 through 24 can be made to be nontransitive. With adjacent pairs, one die will win approximately 2 out of 3 times.

For rolling high number, B beats A, C beats B, D beats C, A beats D.

• A: 1, 2, 16, 17, 18, 19
• B: 3, 4, 5, 20, 21, 22
• C: 6, 7, 8, 9, 23, 24
• D: 10, 11, 12, 13, 14, 15

#### Relation to Efron's dice

These dice are basically the same as Efron's dice, as each number of a series of successive numbers on a single die can all be replaced by the lowest number of the series and afterwards renumbering them.

• A: 1, 2, 16, 17, 18, 19 -> 1, 1, 16, 16, 16, 16 -> 0, 0, 4, 4, 4, 4
• B: 3, 4, 5, 20, 21, 22 -> 3, 3, 3, 20, 20, 20 -> 1, 1, 1, 5, 5, 5
• C: 6, 7, 8, 9, 23, 24 -> 6, 6, 6, 6, 23, 23 -> 2, 2, 2, 2, 6, 6
• D: 10, 11, 12, 13, 14, 15 -> 10, 10, 10, 10, 10, 10 -> 3, 3, 3, 3, 3, 3

### Miwin's dice

Miwins dice
Main article: Miwins dice

Miwin's Dice were invented in 1975 by the physicist Michael Winkelmann.

Consider a set of three dice, III, IV and V such that

• die III has sides 1, 2, 5, 6, 7, 9
• die IV has sides 1, 3, 4, 5, 8, 9
• die V has sides 2, 3, 4, 6, 7, 8

Then:

• the probability that III rolls a higher number than IV is 17/36
• the probability that IV rolls a higher number than V is 17/36
• the probability that V rolls a higher number than III is 17/36

### Three-dice set with minimal alterations to standard dice

The following intransitive dice have only a few differences compared to 1 through 6 standard dice:

• as with standard dice, the total number of pips is always 21
• as with standard dice, the sides only carry pip numbers between 1 and 6
• faces with the same number of pips occur a maximum of twice per die
• only two sides on each die have numbers different from standard dice:
• A: 1, 1, 3, 5, 5, 6
• B: 2, 3, 3, 4, 4, 5
• C: 1, 2, 2, 4, 6, 6

Like Miwin’s set, the probability of A winning versus B (or B vs. C, C vs. A) is 17/36. The probability of a draw, however, is 4/36, so that only 15 out of 36 rolls lose. So the overall winning expectation is higher.

## Freivalds's investigation

The set of nontransitive dice were investigated by the Latvian computer scientist and mathematician Rusins Freivalds. He showed that if there is a set of n dice, and each die beats the next with probability p, then p can be arbitrarily close (but not equal) to 3/4 = 0.75 when n goes to infinity.[citation needed]

## Warren Buffett

Warren Buffett is known to be a fan of nontransitive dice. In the book Fortune's Formula: The Untold Story of the Scientific Betting System that Beat the Casinos and Wall Street, a discussion between him and Edward Thorp is described. Buffett and Thorp discussed their shared interest in nontransitive dice. "These are a mathematical curiosity, a type of 'trick' dice that confound most people's ideas about probability."

Buffett once attempted to win a game of dice with Bill Gates using nontransitive dice. "Buffett suggested that each of them choose one of the dice, then discard the other two. They would bet on who would roll the highest number most often. Buffett offered to let Gates pick his die first. This suggestion instantly aroused Gates's curiosity. He asked to examine the dice, after which he demanded that Buffett choose first."[1]

In 2010, Wall Street Journal magazine quoted Sharon Osberg, Buffett's bridge partner, saying that when she first visited his office 20 years earlier, he tricked her into playing a game with nontransitive dice that could not be won and "thought it was hilarious".[2]

## Nontransitive dice set for three players

M. Oskar van Deventer introduced a set of seven dice (all faces with probability 1/6) as follows:[3]

• A: 2, 2, 14, 14, 17, 17
• B: 7, 7, 10, 10, 16, 16
• C: 5, 5, 13, 13, 15, 15
• D: 3, 3, 9, 9, 21, 21
• E: 1, 1, 12, 12, 20, 20
• F: 6, 6, 8, 8, 19, 19
• G: 4, 4, 11, 11, 18, 18

One can verify that A beats B,C,E; B beats C,D,F; C beats D,E,G; D beats A,E,F; E beats B,F,G; F beats A,C,G; G beats A,B,D. Consequently, for arbitrarily chosen two dice there is a third one that beats both of them. Namely,

• G beats A,B; F beats A,C; G beats A,D; D beats A,E; D beats A,F; F beats A,G;
• A beats B,C; G beats B,D; A beats B,E; E beats B,F; E beats B,G;
• B beats C,D; A beats C,E; B beats C,F; F beats C,G;
• C beats D,E; B beats D,F; C beats D,G;
• D beats E,F; C beats E,G;
• E beats F,G.

Whatever the two opponents choose, the third player will find one of the remaining dice that beats both opponents' dice.

## Nontransitive dodecahedrons

In analogy to the nontransitive six-sided dice, there are also dodecahedrons which serve as nontransitive twelve-sided dice. The points on each of the dice result in the sum of 114. There are no repetitive numbers on each of the dodecahedrons.

The miwin’s dodecahedrons (set 1) win cyclically against each other in a ratio of 35:34.

The miwin’s dodecahedrons (set 2) win cyclically against each other in a ratio of 71:67.

Set 1:

 D III with blue dots 1 2 5 6 7 9 10 11 14 15 16 18 D IV with red dots 1 3 4 5 8 9 10 12 13 14 17 18 D V with black dots 2 3 4 6 7 8 11 12 13 15 16 17
 nontransitive dodecahedron D III nontransitive dodecahedron D IV nontransitive dodecahedron D V

Set 2:

 D VI with yellow dots 1 2 3 4 9 10 11 12 13 14 17 18 D VII with white dots 1 2 5 6 7 8 9 10 15 16 17 18 D VIII with green dots 3 4 5 6 7 8 11 12 13 14 15 16
 nontransitive dodecahedron D VI nontransitive dodecahedron D VII nontransitive dodecahedron D VIII

### Nontransitive prime-numbers-dodecahedrons

It is also possible to construct sets of nontransitive dodecahedrons such that there are no repeated numbers and all numbers are primes. Miwin’s nontransitive prime-numbers-dodecahedrons win cyclically against each other in a ratio of 35:34.

Set 1: The numbers add up to 564.

 PD 11 with blue numbers 13 17 29 31 37 43 47 53 67 71 73 83 PD 12 with red numbers 13 19 23 29 41 43 47 59 61 67 79 83 PD 13 with black numbers 17 19 23 31 37 41 53 59 61 71 73 79
 nontransitive prime-numbers-dodecahedron PD 11 nontransitive prime-numbers-dodecahedron PD 12 nontransitive prime-numbers-dodecahedron PD 13

Set 2: The numbers add up to 468.

 PD 1 with yellow numbers 7 11 19 23 29 37 43 47 53 61 67 71 PD 2 with white numbers 7 13 17 19 31 37 41 43 59 61 67 73 PD 3 with green numbers 11 13 17 23 29 31 41 47 53 59 71 73
 nontransitive prime-numbers-dodecahedron PD 1 nontransitive prime-numbers-dodecahedron PD 2 nontransitive prime-numbers-dodecahedron PD 3